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Sagot :
Let's consider the given short-run production function:
[tex]\[ Q = 6L^2 - 0.4L^3 \][/tex]
We'll address each part of the question step-by-step.
### A. Find the value of [tex]\( L \)[/tex] that maximizes output
To find the value of [tex]\( L \)[/tex] that maximizes the production function [tex]\( Q \)[/tex], we need to find the critical points by setting the first derivative of [tex]\( Q \)[/tex] with respect to [tex]\( L \)[/tex] to zero. The first derivative of [tex]\( Q \)[/tex] is:
[tex]\[ \frac{dQ}{dL} = 12L - 1.2L^2 \][/tex]
Setting this equal to zero to find the critical points:
[tex]\[ 12L - 1.2L^2 = 0 \][/tex]
[tex]\[ 12L = 1.2L^2 \][/tex]
[tex]\[ 12 = 1.2L \][/tex]
[tex]\[ L = 10 \][/tex]
From this, we can conclude that [tex]\( L = 10 \)[/tex] maximizes the output [tex]\( Q \)[/tex].
### B. Find the value of [tex]\( L \)[/tex] that maximizes marginal product
The marginal product (MP) is the derivative of the production function with respect to [tex]\( L \)[/tex]. We already have the marginal product function:
[tex]\[ MP = 12L - 1.2L^2 \][/tex]
To find the value of [tex]\( L \)[/tex] that maximizes the marginal product, we must find the critical points of the marginal product function by taking its derivative and setting it to zero:
[tex]\[ \frac{d(MP)}{dL} = 12 - 2.4L \][/tex]
Setting this equal to zero:
[tex]\[ 12 - 2.4L = 0 \][/tex]
[tex]\[ 2.4L = 12 \][/tex]
[tex]\[ L = 5 \][/tex]
Thus, [tex]\( L = 5 \)[/tex] maximizes the marginal product.
### C. Find the value of [tex]\( L \)[/tex] that maximizes average product
The average product (AP) is the production function [tex]\( Q \)[/tex] divided by [tex]\( L \)[/tex]:
[tex]\[ AP = \frac{Q}{L} = \frac{6L^2 - 0.4L^3}{L} = 6L - 0.4L^2 \][/tex]
To maximize the average product, we take the derivative of the average product function with respect to [tex]\( L \)[/tex] and set it to zero:
[tex]\[ \frac{d(AP)}{dL} = 6 - 0.8L \][/tex]
Setting this equal to zero:
[tex]\[ 6 - 0.8L = 0 \][/tex]
[tex]\[ 0.8L = 6 \][/tex]
[tex]\[ L = 7.5 \][/tex]
Therefore, [tex]\( L = 7.5 \)[/tex] maximizes the average product.
### Summary:
- The value of [tex]\( L \)[/tex] that maximizes output is [tex]\( \boxed{10} \)[/tex].
- The value of [tex]\( L \)[/tex] that maximizes the marginal product is [tex]\( \boxed{5} \)[/tex].
- The value of [tex]\( L \)[/tex] that maximizes the average product is [tex]\( \boxed{7.5} \)[/tex].
[tex]\[ Q = 6L^2 - 0.4L^3 \][/tex]
We'll address each part of the question step-by-step.
### A. Find the value of [tex]\( L \)[/tex] that maximizes output
To find the value of [tex]\( L \)[/tex] that maximizes the production function [tex]\( Q \)[/tex], we need to find the critical points by setting the first derivative of [tex]\( Q \)[/tex] with respect to [tex]\( L \)[/tex] to zero. The first derivative of [tex]\( Q \)[/tex] is:
[tex]\[ \frac{dQ}{dL} = 12L - 1.2L^2 \][/tex]
Setting this equal to zero to find the critical points:
[tex]\[ 12L - 1.2L^2 = 0 \][/tex]
[tex]\[ 12L = 1.2L^2 \][/tex]
[tex]\[ 12 = 1.2L \][/tex]
[tex]\[ L = 10 \][/tex]
From this, we can conclude that [tex]\( L = 10 \)[/tex] maximizes the output [tex]\( Q \)[/tex].
### B. Find the value of [tex]\( L \)[/tex] that maximizes marginal product
The marginal product (MP) is the derivative of the production function with respect to [tex]\( L \)[/tex]. We already have the marginal product function:
[tex]\[ MP = 12L - 1.2L^2 \][/tex]
To find the value of [tex]\( L \)[/tex] that maximizes the marginal product, we must find the critical points of the marginal product function by taking its derivative and setting it to zero:
[tex]\[ \frac{d(MP)}{dL} = 12 - 2.4L \][/tex]
Setting this equal to zero:
[tex]\[ 12 - 2.4L = 0 \][/tex]
[tex]\[ 2.4L = 12 \][/tex]
[tex]\[ L = 5 \][/tex]
Thus, [tex]\( L = 5 \)[/tex] maximizes the marginal product.
### C. Find the value of [tex]\( L \)[/tex] that maximizes average product
The average product (AP) is the production function [tex]\( Q \)[/tex] divided by [tex]\( L \)[/tex]:
[tex]\[ AP = \frac{Q}{L} = \frac{6L^2 - 0.4L^3}{L} = 6L - 0.4L^2 \][/tex]
To maximize the average product, we take the derivative of the average product function with respect to [tex]\( L \)[/tex] and set it to zero:
[tex]\[ \frac{d(AP)}{dL} = 6 - 0.8L \][/tex]
Setting this equal to zero:
[tex]\[ 6 - 0.8L = 0 \][/tex]
[tex]\[ 0.8L = 6 \][/tex]
[tex]\[ L = 7.5 \][/tex]
Therefore, [tex]\( L = 7.5 \)[/tex] maximizes the average product.
### Summary:
- The value of [tex]\( L \)[/tex] that maximizes output is [tex]\( \boxed{10} \)[/tex].
- The value of [tex]\( L \)[/tex] that maximizes the marginal product is [tex]\( \boxed{5} \)[/tex].
- The value of [tex]\( L \)[/tex] that maximizes the average product is [tex]\( \boxed{7.5} \)[/tex].
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